The answer in itself is 1/128 and here is the procedure to prove it:
cos(A)*cos(60+A)*cos(60-A) = cos(A)*(cos²60 - sin²A)
<span>= cos(A)*{(1/4) - 1 + cos²A} = cos(A)*(cos²A - 3/4) </span>
<span>= (1/4){4cos^3(A) - 3cos(A)} = (1/4)*cos(3A) </span>
Now we group applying what we see above
<span>cos(12)*cos(48)*cos(72) = </span>
<span>=cos(12)*cos(60-12)*cos(60+12) = (1/4)cos(36) </span>
<span>Similarly, cos(24)*cos(36)*cos(84) = (1/4)cos(72) </span>
<span>Now the given expression is: </span>
<span>= (1/4)cos(36)*(1/4)*cos(72)*cos(60) = </span>
<span>= (1/16)*(1/2)*{(√5 + 1)/4}*{(√5 - 1)/4} [cos(60) = 1/2; </span>
<span>cos(36) = (√5 + 1)/4 and cos(72) = cos(90-18) = </span>
<span>= sin(18) = (√5 - 1)/4] </span>
<span>And we seimplify it and it goes: (1/512)*(5-1) = 1/128</span>
Answer:
27
Step-by-step explanation:
It should be A= 1/2bh but im not sure why you wrote those extra things. Are we supposed to use them?
Answer:
AAS, the two equal angles, the two right triangles, and the shared side
Step-by-step explanation:
The third one is the correct graph for the equation
hope it helps !!
